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ELEN [110]
3 years ago
12

A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro

duces an upright image exactly 40% the size of the object. What is the distance between the object and its image?
Physics
1 answer:
tekilochka [14]3 years ago
6 0

Answer:

distance between object and image =  18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

distance between object and image

solution

we know lens formula that is

1/f = 1/v + 1/u     ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm   ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm     ......................3

so from equation 2 and 3

distance between object and image =  5.4 + 13.5

distance between object and image =  18.9 cm

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the masses of baseball, basketball,  tennis ball and bowling ball can in arranged in increasing order as

tennis ball < baseball < basketball < bowling ball.

since acceleration and mass have inverse relation from the formula , the order of the acceleration will also be reverse.

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Permanent magnet. An induced magnet would be created when a piece of iron (for example) is in contact with a magnet. Temporary magnets would be something like an electromagnet. Bar magnets are permanently magnetic unless we heat them or hammer them to cause their domains to loose alignment.
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2 years ago
In which parts of a plant would u expect phototropism to occur?
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Answer:

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The time constant for RC circuit with the values of R1 and C1 is 5ms. What will be the time constant for a new RC circuit with t
lions [1.4K]

Answer:

d. 25 ms

Explanation:

  • In a RC circuit we call time constant to the product of the resistance times the capacitance, which represents the time when the charge reaches to the 63% of the final value, as follows:

       \tau_{1} = R_{1} *C_{1}  = 5 ms (1)

  • If we have a new circuit with new values for R and C, the time constant will be defined in the same way, as follows:

       \tau_{2} =10* R_{1} *0.5*C_{1}  = 5*(R_{1}* C_{1}) = 5* \tau_{1} = 5* 5 ms = 25 ms (2)

6 0
2 years ago
A fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predete
zmey [24]

Answer:

The diameter of wire should be 4.04 \times 10^{-4} m

Explanation:

Given:

Current density J = 500 \times 10^{4} \frac{A}{m^{2} }

Current I = 0.64 A

From the formula of current density,

  J = \frac{I}{A}

Where A = area of cylindrical wire = \pi r^{2}

  \pi r^{2} = \frac{I}{J}

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   r = \sqrt{\frac{0.64}{3.14 \times 500 \times 10^{4} } }

   r = 2.02 \times 10^{-4}m

For finding the diameter of wire,

   d = 2r

   d = 4.04 \times 10^{-4}m

Therefore, the diameter of wire should be 4.04 \times 10^{-4} m

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